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According to my book, a decrease in volume will result in an increase in pressure for an equilibrium mixture, which will in turn cause its equilibrium to shift to the side with the least number of gas molecules, given the reaction involves gases (and according to Le Chatelier's Principle).

It also says that Volume and Pressure changes will cause the position of equilibrium to shift, but the equilibrium constant remains unchanged...

If the equilibrium constant (K) depends on the concentrations of reactants and products...

and according to the molarity formula, Volume is inversely proportional with respect to concentration...

... then why does the equilbrium constant remain unchanged?

So far I have this, but I don't know what it means...

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  • $\begingroup$ Surely you have the details confused. Is the book talking about a specific example rather than generalities? $\endgroup$ – MaxW Feb 6 '17 at 19:51
  • $\begingroup$ It's referring to it in general terms. This is exactly what the book says: "A decrease in pressure will cause a shift in the equilibrium position to the side with the larger number of molecules. A different equilibrium position will be achieved, but the value of K will be unchanged, as long as the temperature remains the same" $\endgroup$ – Sam202 Feb 6 '17 at 19:55
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In your workyou have changed the "equilibrium constant" which is a fixed while keeping the number of moles constant of all species in both instances. Rather its the opposite

Consider the reaction;$$\ce{AB <=> A + B }$$ with equilibrium constant $K_c$ at temperature $T$ and volume $V_0$

Initially let $n(AB) = x_0$ , $n(A) = y_0$ , $n(B) = z_0$ where $n()$ refers to the number of moles.

then $$K_c = {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}}$$

Now the volume is reduuced from $\ce{V_0\bond{->}V}$ Hence all the concentrations increase. Then the reaction quotient at the moment the volume is changed;$$Q_c = {\frac{{\frac {x_0}{V}}{\frac {y_0}{V}}}{{\frac {z_0}{V}}}}$$

Now $Q_c > K_c$ hence equilibrium shifts to the right reducing the number of moles of $A$ and $B$ increasing the number of moles of $AB$.Due to this effect $Q_c$ will gradually reduce and at acertain point will becaome equal tp $K_c$ with different number of moles.

$$K_c = {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}} = {\frac{{\frac {x}{V}}{\frac {y}{V}}}{{\frac {z}{V}}}} $$

The only parameter remaining constant(under constant temperature) is the Equilibrium constant is a certain parameter(i.e. Volume here) is changed, other parameters will vary keeping Equilibrium constant a constant.

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"A decrease in pressure will cause a shift in the equilibrium position to the side with the larger number of molecules. A different equilibrium position will be achieved, but the value of K will be unchanged, as long as the temperature remains the same"

The book has overgeneralized. Look at this reaction:

$\ce{2BrCl(g) <=> Br2(g) + Cl2(g)}$

There are the same number of molecules on both sides. So decreasing the pressure doesn't always cause a shift in equilibrium.

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    $\begingroup$ I found the problem. I was confusing the reaction quotient Q with the equilibrium constant K. Q changes with all sorts of factors like pressure, volume, catalysts, and temperature, but K only changes with temperature. According to other comments, Q eventually becomes equal to K again after some time when the equilibrium is re-established. $\endgroup$ – Sam202 Feb 6 '17 at 20:21
  • $\begingroup$ Q isn't the problem... $\endgroup$ – MaxW Feb 6 '17 at 20:22
  • $\begingroup$ I didn't include the whole lecture above, but it does mention what you said. If the change in gas molecules between reactants and products is 0, then a change in pressure or volume will not change the equilibrium position or K. However, when the change in gas molecules is not 0, then the equilibrium position will change, but not K. It says K only changes with temperature. $\endgroup$ – Sam202 Feb 6 '17 at 20:28
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    $\begingroup$ Make Cl diatomic. I couldn't do it for you, evidently there need be 6 or more characters in an edit. $\endgroup$ – bpedit Feb 6 '17 at 22:40
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    $\begingroup$ Both the question and your bold quotes refer to examples where one side has more moles of gas. Your example doesn't appear to apply. $\endgroup$ – bpedit Feb 6 '17 at 22:44
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Le Chatelier talks only about how the and on which side will the equilibrium position shift. Not about the change in equilibrium constant. Let me make two points. :-

  • Change in volume affecting the equilibrium constant:- $\Delta$n = (sum of stoichiometric coefficient of products)-(sum of stoichiometric coefficient of products). E.G.

$$\ce{aA + bB <=> cC + dD}$$

here $\Delta$n= (c+d)-(a+b)

now,

$$K_c=\frac{[C]^c [D]^d}{[A]^a[B]^b}$$

since $$[A]=\frac{moles_a}{volume(V)}$$

similarly for all . $$K_c=K_x*V^{\Delta n}$$

So if

$\Delta n= 0$ then it doesn't affect

$\Delta n >0$ the it is directly proportional

$\Delta n <0$ the it is inversely proportional

E.G.

$$\ce{H_2 + I_2 <=> 2HI} \Delta n= 0$$

$$\ce{N_2 + 3H_2 <=> 2NH_3} \Delta n= (-2)$$

$$\ce{ PCl_3 <=> PCl_5 + Cl_2} \Delta n= (1)$$

  • Le Chatelier says that if volume is increased then number of moles per volume decreases then the equilibrium will shift towards the side with more moles which is evident from the previous explanation.

If you still find a doubt then let me know I will like to give another try with more explanation

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